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JEE Main 202127 JulyEvening Shift
Exam
JEE Main
Year
2021
Date
27 July
Shift
Evening Shift
Subject
Physics
Chapter
Motion in One Dimension

JEE Main 2021 · 27 July · Evening Shift · Physics · Motion in One Dimension

A particle of mass M originally at rest is subjected to a force...

A particle of mass MM originally at rest is subjected to a force whose direction is constant but magnitude varies with time according to the relation F=F0F = {F}_{0}\left[1 - {\left(\dfrac{t - T}{T}\right)}^{2}\right] where F0{F}_{0} and TT are constants. The force acts only for the time interval 2T2T. The velocity vv of the particle after time 2T2T is:

Options

  1. A2F0TM\dfrac{2{F}_{0}T}{M}
  2. BF0T2M\dfrac{{F}_{0}T}{2M}
  3. C4F0T3M\dfrac{4{F}_{0}T}{3M}
  4. DF0T3M\dfrac{{F}_{0}T}{3M}

Correct answer

C. 4F0T3M\dfrac{4{F}_{0}T}{3M}

Step-by-step solution

t=0,u=0t = 0, u = 0

a=FoM−FoMT2(t−T)2=dvdta = \dfrac{{F}_{o}}{M} - \dfrac{{F}_{o}}{M{T}^{2}} ( t - T{)}^{2} = \dfrac{dv}{dt}

∫0vdv=∫t=02T(FoM−FoMT2(t−T)2)dt{∫}_{0}^{v}dv = {∫}_{t = 0}^{2T}\left(\dfrac{{F}_{o}}{M} - \dfrac{{F}_{o}}{M{T}^{2}} ( t - T{)}^{2}\right)dt

V=[FoMt]o2T−FoMT2[t33−t2T+T2t]02TV = {\left[\dfrac{{F}_{o}}{M}t\right]}_{o}^{2T} - \dfrac{{F}_{o}}{M{T}^{2}}{\left[\dfrac{{t}^{3}}{3} - {t}^{2}T + {T}^{2}t\right]}_{0}^{2T}

V=4FoT3MV = \dfrac{4{F}_{o}T}{3M}

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